Lemma 18.26.16.4C3ACB006A19 Use the Green potential $g$, the probability measure $\sigma $ and the sets $D=\Delta ^ n$, $S=K^ n$ of lemma 18.26.15. For $s,L>0$ and $a=(a_1,\ldots ,a_ n)\in (0,+\infty )^ n$, put
\[ \mu _ s=\mathrm{dd}^{\mathrm{c}}\max (g,-s),\qquad h_ a(z)=\max _ j a_ jg(z_ j), \]
and $h_{a,L}=\max (h_ a,-L)$. Each $\mu _ s$ is a nonatomic probability measure supported on $\{ g=-s\} $, and $\mu _ s\to \sigma $ weakly as $s\to \infty $. The bounded products satisfy
\[ (\mathrm{dd}^{\mathrm{c}}h_{a,L})^ n =\left(\prod _{j=1}^ n a_ j\right) \mu _{L/a_1}\otimes \cdots \otimes \mu _{L/a_ n}. \]
The function $h_ a$ has pole set $S$, is locally Lipschitz off $S$, and has zero bounded top-degree product there. Its full product in the sense of theorem 5.27.18 is
\[ (\mathrm{dd}^{\mathrm{c}}h_ a)^ n_{\mathrm{full}} =\left(\prod _{j=1}^ n a_ j\right)\sigma ^{\otimes n}. \]
In particular, the common mass $M$ in lemma 18.26.15 is $1$.
Proof.
The truncation $\max (g,-s)$ agrees with $g$ near the boundary of $\Delta $. Their difference is a compactly supported integrable function, so its complex Hessian has integral zero, and $\mu _ s(\Delta )=\sigma (\Delta )=1$. The truncation is constant near $K$ and harmonic off $K$ wherever $g\ne -s$, which proves the support assertion. Since the truncation is bounded, $\mu _ s$ has no point masses, by theorem 5.27.1. Decreasing local $L^1$ convergence to $g$ gives $\mu _ s\to \mathrm{dd}^{\mathrm{c}}g=\sigma $; for large $s$, the supports lie in one fixed compact subset of $\Delta $.
The rank argument of lemma 18.26.15 applies with the weights $a_ j$. Away from $S$, the finite branches dominate the branches near poles, so $h_ a$ is locally a Lipschitz maximum of at most $n$ pluriharmonic functions, with zero top-degree product. For the bounded maximum with the constant branch $-L$, the same argument shows that its top-degree product is supported on
\[ Q_{a,L}=\{ z:a_ jg(z_ j)=-L\text{ for every }j\} . \]
Indeed, off this set one branch lies strictly below the maximum, leaving at most $n$ active pluriharmonic branches; this also applies near branches with poles, which lie below the constant branch.
Near a point of $Q_{a,L}$ at which all the differentials $\mathrm dg(z_ j)$ are nonzero, set $x_ j=a_ jg(z_ j)+L$. Each $x_ j$ is the real part of a local holomorphic coordinate in its own factor. For $\varepsilon >0$, smooth the maximum locally by
\[ f_\varepsilon (x)=\varepsilon \log \left(1+\sum _{j=1}^ n e^{x_ j/\varepsilon }\right), \]
with weights
\[ w_0=\frac1{1+\sum _ j e^{x_ j/\varepsilon }},\qquad w_ j=e^{x_ j/\varepsilon }w_0. \]
Then $-L+f_\varepsilon $ decreases to $h_{a,L}$. The Hessian matrix of $f_\varepsilon $ in the variables $x_ j$ is $\varepsilon ^{-1}(\operatorname {diag}(w_ j)-(w_ jw_ k)_{j,k})$, whose determinant is $\varepsilon ^{-n}w_0\prod _ jw_ j$. Hence
\[ (\mathrm{dd}^{\mathrm{c}}f_\varepsilon )^ n =n!\varepsilon ^{-n}w_0\prod _{j=1}^ n w_ j \bigwedge _{j=1}^ n(\mathrm dx_ j\wedge \mathrm d^{\mathrm c}x_ j). \]
The scalar kernel on $\mathbb R^ n$ is an approximate identity of mass $1$. To check the constant, write $y=x/\varepsilon $. The map
\[ y\longmapsto \left(\frac{e^{y_ j}}{1+\sum _ k e^{y_ k}}\right)_{j=1}^ n \]
is a diffeomorphism onto the open simplex $\{ w_ j>0,\ \sum _ jw_ j<1\} $, with Jacobian $w_0\prod _ jw_ j$. The simplex has volume $1/n!$, so the kernel above has integral $1$. Integrability of the fixed kernel in $y$ gives concentration at $x=0$ as $\varepsilon \downarrow 0$.
Bounded decreasing convergence of products therefore identifies the local limit with the tensor product of the one-variable level measures. In the $j$th factor, the chain rule for $\max (x_ j,0)$ gives
\[ \delta _0(x_ j)\, \mathrm dx_ j\wedge \mathrm d^{\mathrm c}x_ j =a_ j\mathrm{dd}^{\mathrm{c}}\max (g,-L/a_ j), \]
where the Dirac notation denotes the level measure in the real normal coordinate $x_ j$. The critical points of the nonconstant harmonic function $g$ off $K$ are isolated. In each coordinate neighborhood, their inverse images are analytic hypersurfaces, hence pluripolar. The bounded top-degree product gives them zero mass by theorem 5.27.1, and so does the candidate tensor measure, because each $\mu _{L/a_ j}$ has no point masses. Thus the formula extends across these points and, together with the support assertion, holds on all of $D$.
Tensor products of the curve probabilities converge to $\sigma ^{\otimes n}$: this follows first for products of continuous functions of one variable, and then for all continuous test functions by uniform approximation on a common compact product. Finally, $h_{a,L}\downarrow h_ a$, whose unbounded locus is the closed set $S$ of Hausdorff dimension zero. By theorem 5.27.18, the weak limit is the full top-degree product of $h_ a$. Taking $a_ j=1$ proves $M=1$.
$\square$
Pome
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